Question What is the particular solution to the differential equation (dy)/(dx)=(1+y)/(x) with the initial condition y(e)=1 ?

Recognize separable form: Step 1: Recognize the differential equation as a separable form. We can write it as dy/(1+y) = dx/x. Integrate both sides: Step 2: Integrate both sides. The left side becomes ∫dy/(1+y) and the right side becomes ∫dx/x. Perform integration: Step 3: Perform the integration. The integral of 1/(1+y) dy is ln|1+y| and the integral of 1/x dx is ln|x|. So, ln|1+y| = ln|x| + C, where C is the integration constant. Solve for y: Step 4: Solve for y. Exponentiate both sides to remove the logarithm, getting |1+y| = e^(ln|x| + C) = |x|e^C. Apply initial condition: Step 5: Apply the initial condition y(e) = 1. Substituting x = e and y = 1 into |1+y| = |x|e^C gives |1+1| = |e|e^C, so 2 = e^C. Solve for C: Step 6: Solve for C. Taking the natural logarithm of both sides, ln(2) = C. Substitute back into equation: Step 7: Substitute C back into the equation for y. We have |1+y| = |x|e^(ln(2)), or |1+y| = 2|x|. Final solution: Step 8: Solve for y. We get 1+y = 2x or y = 2x - 1.

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Question
What is the particular solution to the differential equation
(dy)/(dx)=(1+y)/(x) with the initial condition
y(e)=1 ?

$\newline$ What is the particular solution to the differential equation $\frac{d y}{d x}=\frac{1+y}{x}$ with the initial condition $y(e)=1$ ?

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Algebra 2

Simplify expressions using trigonometric identities

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What is the particular solution to the differential equation

\frac{d y}{d x}=\frac{1+y}{x}

with the initial condition

y(e)=1

Recognize separable form: Step $1$ : Recognize the differential equation as a separable form. We can write it as $\frac{dy}{1+y} = \frac{dx}{x}$ .
Integrate both sides: Step $2$ : Integrate both sides. The left side becomes $\int \frac{dy}{1+y}$ and the right side becomes $\int \frac{dx}{x}$ .
Perform integration: Step $3$ : Perform the integration. The integral of $\frac{1}{1+y} \, dy$ is $\ln|1+y|$ and the integral of $\frac{1}{x} \, dx$ is $\ln|x|$ . So, $\ln|1+y| = \ln|x| + C$ , where $C$ is the integration constant.
Solve for y: Step $4$ : Solve for y. Exponentiate both sides to remove the logarithm, getting $|1+y| = e^{\ln|x| + C} = |x|e^C$ .
Apply initial condition: Step $5$ : Apply the initial condition $y(e) = 1$ . Substituting $x = e$ and $y = 1$ into $|1+y| = |x|e^C$ gives $|1+1| = |e|e^C$ , so $2 = e^C$ .
Solve for C: Step $6$ : Solve for C. Taking the natural logarithm of both sides, $\ln(2) = C$ .
Substitute back into equation: Step $7$ : Substitute $C$ back into the equation for $y$ . We have $|1+y| = |x|e^{\ln(2)}$ , or $|1+y| = 2|x|$ .
Final solution: Step $8$ : Solve for $y$ . We get $1+y = 2x$ or $y = 2x - 1$ .